We are tasked with finding the recursive formula for the coefficients $a_n$ in the power series solution. Let’s work through this step by step.

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Given:

The differential equation after combining summations becomes:

$\sum_{n=0}^\infty \left[ -5(n+2)(n+1)a_{n+2} + (n+1)^2a_{n+1} + (4n + 3)a_n \right]x^n = 0.$

Since the power series equals zero for all $x$, the coefficients of each power of $x^n$ must be zero. Therefore:

$-5(n+2)(n+1)a_{n+2} + (n+1)^2a_{n+1} + (4n + 3)a_n = 0.$

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Rearrange for $a_{n+2}$:

To express $a_{n+2}$ in terms of $a_{n+1}$ and $a_n$, solve for $a_{n+2}$:

$-5(n+2)(n+1)a_{n+2} = -(n+1)^2a_{n+1} - (4n + 3)a_n.$

Divide through by $-5(n+2)(n+1)$:

$a_{n+2} = \frac{(n+1)^2a_{n+1} + (4n + 3)a_n}{5(n+2)(n+1)}.$

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Recursive Formula:

The recursive formula is:

$a_{n+2} = \frac{(n+1)^2}{5(n+2)(n+1)}a_{n+1} + \frac{4n + 3}{5(n+2)(n+1)}a_n.$

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Initial Conditions:

From $y(0) = 3$ and $y'(0) = 4$:

• $a_0 = 3$,
• $a_1 = 4$.

You can now compute higher-order terms iteratively using the recurrence relation.

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Would you like me to compute specific terms ($a_2, a_3, \dots$) or explore further properties of this solution?

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Related Questions:

1. What are the first four terms of the series solution?
2. How does the recurrence relation help determine the behavior of the solution for large $n$?
3. What is the radius of convergence of this power series?
4. How do the initial conditions affect the final solution?
5. How does the solution behave near $x = 0$?

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Tip:

When working with recurrence relations, always double-check the coefficients in front of $a_{n+2}, a_{n+1},$ and $a_n$ for correctness, as small errors propagate during iterative computation.